Problem: Simplify the following expression: $y = \dfrac{-6x^2+17x- 10}{-6x + 5}$
Solution: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-6)}{(-10)} &=& 60 \\ {a} + {b} &=& &=& {17} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $60$ and add them together. The factors that add up to ${17}$ will be your ${a}$ and ${b}$ When ${a}$ is ${5}$ and ${b}$ is ${12}$ $ \begin{eqnarray} {ab} &=& ({5})({12}) &=& 60 \\ {a} + {b} &=& {5} + {12} &=& 17 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-6}x^2 +{5}x) + ({12}x {-10}) $ Factor out the common factors: $ x(-6x + 5) - 2(-6x + 5)$ Now factor out $(-6x + 5)$ $ (-6x + 5)(x - 2)$ The original expression can therefore be written: $ \dfrac{(-6x + 5)(x - 2)}{-6x + 5}$ We are dividing by $-6x + 5$ , so $-6x + 5 \neq 0$ Therefore, $x \neq \frac{5}{6}$ This leaves us with $x - 2; x \neq \frac{5}{6}$.